Thank you Determine if sigma^infinity_n=1 2n/3n + 1 is convergent or divergent C

Question

Thank you Determine if sigma^infinity_n=1 2n/3n + 1 is convergent or divergent Compute the sum of the series sigma^infinty_n=1 (-1)^n=1(2/pi)^n Detennine whether the series is convergent or divergent. Find the sum if it is convergent. 4 + 8/5 + 16/25 + 32/125 + ... 1 - 3/2 + 9/4 - 27/8 + ... sigma^infinity_n=1 1/e^2n sigma^infinity_n=1 3/n sigma^infinity_n=1 1/n(n+2) sigma^infinity_n=1 ln(n/2n+5) Use the integral test to determine which of the following series are convergent. sigma^infinity_n=1 2- root n/ n^4 sigma^infinity_n=1 e^1/n /n^2

Explanation / Answer

1)an=2n/(3n+1)

nth term limit test:

limn->infinity an

=limn->infinity 2n/(3n+1)

=limn->infinity 2n/(n(3+(1/n)))

=limn->infinity 2/(3+(1/n))

=2/(3+(0))

=2/3

not eqqual to zero

so series  n=1 to infinity2n/(3n+1) is divergent

2)

s=(2/pi) -(2/pi)2 +(2/pi)3 -(2/pi)4+....

(2/pi)s=(2/pi)2 -(2/pi)3 +(2/pi)4 -(2/pi)5+....

s +(2/pi)s=(2/pi) -(2/pi)2 +(2/pi)3 -(2/pi)4+....(2/pi)2 -(2/pi)3 +(2/pi)4 -(2/pi)5+....

s +(2/pi)s=(2/pi)

s(1+ (2/pi))=(2/pi)

s((pi+ 2)/pi)=(2/pi)

s=2/(pi+ 2)

n=1 to infinity (-1)n-1(2/pi)n =2/(pi+ 2)

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